Don't display if no records exist in mysql - php

Question

I have this form to search names in mysql database
<form action="search.php" method="GET">
<input type="text" placeholder="Search" name="name">
<input type="submit" value="Search">
this is the search.php
<?php
name = $_GET['name'];
require_once("connect.php");
$records = $connect->query("SELECT * FROM Userlists WHERE Name = '$name'");
echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Description</th>
</tr>
</thead>
<tbody>";
if (mysqli_num_rows($records)== 0){
echo "No data available for that name specified";
}
else {
while($row=mysqli_fetch_array($records)) {
$name = $row['Name'];
$email = $row['Email'];
$desc = $row['Desc'];
echo
"<tr>
<td>".$name."</td>
<td>".$email."</td>
<td>".$desc."</td>
</tr>";
}
}
echo
"</tbody>
</table>";
?>
so there is no problems when I search for a name that exists in database it displays correctly, but the problem comes when I search for a name that doesn't exist in database.. I want it to display only
"No data available for that name specified" for the output but I will also see empty table in the output like this ------------> IMAGE..
so how can I get rid of the empty table for the output?

Answer 1

Just pu the if outside the table....
<?php
name = $_GET['name'];
require_once("connect.php");
$records = $connect->query("SELECT * FROM Userlists WHERE Name = '$name'");
if (mysqli_num_rows($records)== 0){
echo "No data available for that name specified";
} else {
echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Desc</th>
</tr>
</thead>
<tbody>";
while($row=mysqli_fetch_array($records)) {
$name = $row['Name'];
$email = $row['Email'];
$desc = $row['Desc'];
echo
"<tr>
<td>".$name."</td>
<td>".$email."</td>
<td>".$desc."</td>
</tr>";
}
echo
"</tbody>
</table>";
}
?>

Answer 2

change your if clause as below and remember to add exit() or die() function,this will end your php if there is no any data in database, and if there is any it will then start creating table for once and repeatedly fill up the table rows for given rows of data on database.
if (mysqli_num_rows($records)== 0){
echo "No data available for that name specified";
exit();
} else {
echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Desc</th>
</tr>
</thead>
<tbody>";
while($row=mysqli_fetch_array($records)) {
$name = $row['Name'];
$email = $row['Email'];
$desc = $row['Desc'];
echo
"<tr>
<td>".$name."</td>
<td>".$email."</td>
<td>".$desc."</td>
</tr>";
}
echo
"</tbody>
</table>";
}

Answer 3

Move your echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Desc</th>
</tr>
</thead>
<tbody>";
into the the if statement. This way it will only display the table when data is available!

Answer 4

$name = $_GET['name'];
require_once("connect.php");
$records = $connect->query("SELECT * FROM Userlists WHERE Name = '$name'");
if (mysqli_num_rows($records)== 0){
echo "No data available for that name specified";
}
else {
echo
"<table>
<thead>
<tr>
<th>Name</th>
<th>Email</th>
<th>Desc</th>
</tr>
</thead>
<tbody>";
while ($row = mysqli_fetch_array($records)) {
$name = $row['Name'];
$email = $row['Email'];
$desc = $row['Desc'];
echo
"<tr>
<td>" . $name . "</td>
<td>" . $email . "</td>
<td>" . $desc . "</td>
</tr>";
}
echo "</tbody></table>";
}
Try this one. But don't forget to escape $_GET['name'] like htmlspecialchars and real_escape_string